Electric current definition

Question

I'm trying to understand electric current. Some resources say that it is the flow of charge, and other resources say that it is the quantity of charge that passes through a cross-sectional area over a period of time. This confuses me because I'm not sure if it is a quantity (quantity of charge) or an action (the flow of charge). Can you please provide me with a definitive definition?

Answer 1

Simply stated, current is just a flow of charge. If, however you want to measure and quantify the amount of current, the quantity of current is the amount of charge passing a point over a period of time.

The first statement defines current, while the second defines its measurement.

Answer 2

You can probably find various definitions so I am not sure if there is any single "definitive" definition of electric current.

But as an engineer, I use the definition from the reference handbook for the Fundamentals Exam for becoming a professional engineering in the U.S. as follows:

"Electric current $i(t)$ through a surface is defined as the rate of charge transport through that surface or

$$ i(t)=\frac{dq(t)}{dt}$$

which is a function of time $t$, since $q(t)$ denotes instantaneous charge.

A constant current $i(t)$ is written as $I$, and the vector current density in amperes/m$^2$ is defined as $\vec J$."

Hope this helps.

Answer 3

Your confussion regarding the terms is specific to the use of the terms in English. There is the phenomenon of charge flowing through a wire and the physical quantity that is used to measure this flow. The two are different objects and they could be named with two different words. Which is done in french, Italian, Romanian and maybe other languages that I don't know. So the phenomenon is the "electrical current" and the quantity representing the rate of flow of charge is the "intensity of the electrical current". A hint about this is the usual leter used to desigante it "I" and not "C" (from current). The "intensity of the current" can be shorted to just "current" and so it may produce confusion with the term used for the phenomenon. In English this is so common that even in physics textbooks they call the quantity "current" even though they mean the intensity of the current. In Romanian, for example, I learned about the "intensity of the current" in school and this is what we called it in the classroom, but people in electrical industries and trades call it "current" as well or even the equivalent of amperage, like in English we use voltage for potential difference. In French and Italian physics texts I can see that they also use the equivalent of "intensity of the current" so there is less confusion than in English, I suppose.

Answer 4

Bob D and SomeUser, respectively, explained that current is the movement of charge per unit time, $$ I = \frac{\partial Q_\Omega}{\partial t},$$ and the integral of current density over cross sectional area, $$ I = \int_{\partial\Omega} \text{d}\mathbf{S} \cdot \mathbf{J}.$$ I will add that these two quantities are the same because of charge conservation. Indeed, by standard arguments (see Griffiths, Electrodynamics), we can write down the following equation $$\frac{\partial \rho}{\partial t} + \nabla\cdot \mathbf{J} = 0.\qquad\qquad(*)$$ The first term describes the change of charge density over time in some region $\Omega$, and the second describes the outward flow of the current density in that same region. Indeed, we see the quantities above once we integrate over the region $\Omega$, $$\int \text{d}V ~\frac{\partial \rho}{\partial t} = \frac{\partial Q}{\partial t}, \quad\text{and} \quad \int_\Omega \text{d}V~ \nabla \cdot \mathbf{J} = \int_{\partial\Omega}\text{d}\mathbf{S}\cdot \mathbf{J},$$ where we've used the divergence theorem for the second equality.

Since charge density and current density are not in general separately conserved, we can define a new quantity, called the four-current, which is conserved by the equation $(*)$: $$J^\mu = (\rho, \mathbf{J}),\qquad \partial_\mu J^\mu = \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0.$$ The statement $\partial_\mu J^\mu= 0$ is equivalent to the conservation of charge. See here for more.

Answer 5

Electric current $I$ through a cross section of a conductor is: $$\lim_{\Delta t \to 0} \frac{\Delta q}{\Delta t}$$ where $\Delta q$ is the amount of charge that passed through that cross section during the period of time of length $\Delta t$.

Therefore, it is a scalar quantity.

Alternatively, if the area of the cross section is $S$, the quantity of charge per unit volume of the conductor is $n=\frac{N}{V}$ ($N$ is the number of charged particles in volume $V$) and $v_d$ the average drift velocity in the direction of the electric field, then the amount of charge that passes through that cross section in time $\Delta t$ is $\Delta q=q_0 \times n \times S \times v_d \times \Delta t$ ($q_0$ is the elementary charge of a single particle).

This is because all the charge that passes through during this period is located within a cylinder of height $v_d \Delta t$ and volume $V=Sv_d \Delta t$ which contains exactly $\Delta N=nV=nSv_d \Delta t$ charged particles.

From this, the current $I_S$ through this cross section would be $$\lim_{\Delta t \to 0} \frac{\Delta q}{\Delta t}=\lim_{\Delta t \to 0} \frac{q_0nSv_d \Delta t}{\Delta t}=\lim_{\Delta t \to 0} q_0nSv_d=q_0nSv_d$$.

All in all, it is a scalar quantity that describes the amount of charge that flows through a surface during a certain period. Its units are $\frac{Coulombs}{second} \equiv Amperes$.

Answer 6

I prefer to define it as \begin{equation} I = \iint_{S} \mathbf{J} \cdot \, d\mathbf{S} \end{equation}

which can be rewritten as \begin{align} I &= \frac{\iint_{S} \rho \mathbf{v} dt \cdot \, d\mathbf{S}}{dt} \\ \end{align} The charge density is integrated over the surface and some thickness given by the velocity (actually, just the speed on the direction perpendicular to $S$) of the charges: if the charges are moving faster, further charges will reach the surface in the same time $dt$. You can see units match up. \begin{align} &= \frac{\iint_{S} \rho d\mathbf{r} \cdot \, d\mathbf{S}}{dt} \\ &= \frac{\iint_{S} \rho \,dr_{\perp} \,dS}{dt} \\ \end{align} Bassically, the numerator is the total amount of charge that will pass trough $S$ in $dt$. Thus, we can write it $dq$. \begin{equation} I = \frac{dq}{dt} \end{equation}

This do not intend to be a "proof" nor similar. I just wanted to point out that both concecpts are equivalent. Myself, I was introduced this concept the other way arround.